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vichka [17]
3 years ago
5

2. Use the diagram below to answer this question. As the ball moves from point A

Physics
1 answer:
Thepotemich [5.8K]3 years ago
5 0

Answer:

at point A the ball possess pontetial energy , point B kinetic energy then point C pontetial energy

You might be interested in
Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a
mezya [45]

Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

q= +2e

The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

V= 9×10^9×(5.67×10^-14)= 5101.35v

3 0
3 years ago
A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the
Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

 

 Goodluck

Explanation:

4 0
2 years ago
When sound is created it travels to the car through a
Rasek [7]

A sound wave is a longitudinal wave caused by vibrations and carried through a substance. The particles of the substance, such as air particles, vibrate back and forth along the path that the sound waves travel. Sound is transmitted through the vibrations and collisions of the particles.

This could maybe help you with your answer.

4 0
3 years ago
An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit
mylen [45]

Find how much work ∆<em>W</em> is done by the motor in lifting the elevator:

<em>P</em> = ∆<em>W</em> / ∆<em>t</em>

where

• <em>P</em> = 45.0 kW = power provided by the motor

• ∆<em>W</em> = work done

• ∆<em>t</em> = 20.0 s = duration of time

Solve for ∆<em>W</em> :

∆<em>W</em> = <em>P</em> ∆<em>t</em> = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is <em>M</em> = (<em>m</em> + 490.0) kg, where <em>m</em> is the mass of the elevator alone. Then

∆<em>W</em> = <em>M</em> <em>g h</em>

where

• <em>g</em> = 9.80 m/s² = acceleration due to gravity

• <em>h</em> = 35.0 m = distance covered by the elevator

Solve for <em>M</em>, then for <em>m</em> :

<em>M</em> = ∆<em>W</em> / (<em>g h</em>) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

<em>m</em> = <em>M</em> - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0
3 years ago
A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
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